Given equation of circle :

$x^2+y^2+4x-6y=12$

Now equation of tangent to this circle at $\left(1,-1\right)$ is

$T=0\Rightarrow x-y+4 \left(\frac{x+1}{2}\right)-6 \left(\frac{y-1}{2}\right)=12$

$\Rightarrow x-y+2x+2-3y+3=12$

$\Rightarrow 3x-4y-7=0$

Now equation of circle touching the line $3x-4y-7=0$ at $\left(1, -1\right)$ is

${\left(x-1\right)}^2+{\left(y+1\right)}^2+\lambda \left(3x-4y-7\right)=0 ...\left(i\right)$

$\because$ It passes through $\left(4, 0\right)$

$$\begin{gathered} \Rightarrow {\left(4-1\right)}^2+{\left(0+1\right)}^2+\lambda \left(3\times 4-4\times 0-7\right)=0 \\ \Rightarrow \lambda =-2 \end{gathered}$$

Put $\lambda =-2$ in (i) we get equation of required circle

$\Rightarrow {\left(x-1\right)}^2+{\left(y+1\right)}^2-2 \left(3x-4y-7\right)=0$

$\Rightarrow x^2+y^2-8x+10y+16=0$

$\Rightarrow$Radius$=\sqrt{{\left(-4\right)}^2+{\left(5\right)}^2-16}=5$